Q112 / 180 · COMEDK 2015 · Mathematics
If $A=\left[\begin{array}{ll}1 & 0 \\ 1 & 1\end{array}\right]$ and $A^{8}=a A+b I$, then $(a, b)=$
- A$(8,7)$
- B$(-7,8)$
- Correct answerC$(8,-7)$
- D$(-8,-7)$
Solution
We have, $A=\left[\begin{array}{ll}1 & 0 \\ 1 & 1\end{array}\right]$
$A^{2}=A \cdot A=\left[\begin{array}{ll}1 & 0 \\ 1 & 1\end{array}\right]\left[\begin{array}{ll}1 & 0 \\ 1 & 1\end{array}\right]=\left[\begin{array}{ll}1 & 0 \\ 2 & 1\end{array}\right]$
$A^{3}=A^{2} \cdot A=\left[\begin{array}{ll}1 & 0 \\ 2 & 1\end{array}\right]\left[\begin{array}{ll}1 & 0 \\ 1 & 1\end{array}\right]=\left[\begin{array}{ll}1 & 0 \\ 3 & 1\end{array}\right]$
Similarly, $A^{8}=\left[\begin{array}{ll}1 & 0 \\ 8 & 1\end{array}\right]$
Now, $A^{8}=a A+b I$
$$
\begin{aligned}
&\Rightarrow\left[\begin{array}{ll}
1 & 0 \\
8 & 1
\end{array}\right]=a\left[\begin{array}{ll}
1 & 0 \\
1 & 1
\end{array}\right]+b\left[\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right] \\
&\Rightarrow\left[\begin{array}{ll}
1 & 0 \\
8 & 1
\end{array}\right]=\left[\begin{array}{cc}
a+b & 0 \\
a & a+b
\end{array}\right] \\
&\therefore a=8 \text { and } a+b=1 \\
&\Rightarrow b=1-8=-7
\end{aligned}
$$
Hence, the correct answer is option: 'C'.